Principal Examiner Feedback

Summer 2018 Pearson Edexcel GCE Mathematics

Core Mathematics C4 (6666/01)

Edexcel and BTEC Qualifications

Edexcel and BTEC qualifications are awarded by Pearson, the UK’s largest awarding

body. We provide a wide range of qualifications including academic, vocational,

occupational and specific programmes for employers. For further information visit our

qualifications websites at www.edexcel.com or www.btec.co.uk. Alternatively, you can

get in touch with us using the details on our contact us page at

www.edexcel.com/contactus.

Pearson: helping people progress, everywhere Pearson aspires to be the world’s leading learning company. Our aim is to help

everyone progress in their lives through education. We believe in every kind of

learning, for all kinds of people, wherever they are in the world. We’ve been involved

in education for over 150 years, and by working across 70 countries, in 100 languages,

we have built an international reputation for our commitment to high standards and

raising achievement through innovation in education. Find out more about how we

can help you and your students at: www.pearson.com/uk

Summer 2018

Publications Code 6666_01_1806_ER

All the material in this publication is copyright

© Pearson Education Ltd 2018

Core Mathematics 4 (6666) – Principal Examiner’s report

General Introduction

This paper offered good discrimination between candidates of all abilities and there

were sufficient opportunities for a typical E grade candidate to gain some marks across

all the questions. There were some testing questions involving parametric equations,

differential equations, vectors and integration that allowed the paper to discriminate

well between the higher grades.

In summary, Q1(a), Q2(a), Q3(i), Q4, Q5, Q7(a), Q7(d) and Q8(a) were a good source

of marks for the average candidate, mainly testing standard ideas and techniques

whereas Q1(b), Q2(b), Q3(ii), Q3(iii), Q6, Q7(b), Q7(c) and Q8(b) were discriminating

at the higher grades. Q7(e) proved to be the most challenging question on the paper.

Question 1

Part (a) was accessible with most candidates scoring full marks and part (b) offered

good discrimination for the more able candidates.

In part (a), most candidates manipulated 4 9x to give

1

292 1 ,

4

x

with the 2

outside the brackets sometimes written incorrectly as either 4, 1 or 1

.2

Many

candidates used a correct method for expanding a binomial expansion of the form

(1 ) ,nkx with some applying incorrect 9

4k or

1.

2n Common mistakes in this

part included sign errors, bracketing errors, simplification errors and forgetting to

multiply the correct 29 81

1 ...8 128

x x

by 2.

In part (b), most candidates solved the equation 4 9 310x to give 34.x

Many candidates substituted 34x (or sometimes 34x ) into their part (a) answer

to give an estimate of 1384.563 for 310 , even though the question clearly stated

that the binomial expansion is valid for 4

.9

x Only a minority of candidates

(including some who had previously used 34)x , simplified 310 to give 10 3.1

and deduced that they needed to substitute 0.1x into their binomial expansion. Many

of these candidates arrived at the correct answer 17.623, although a few gave their final

answer as 1.762 by forgetting to multiply their correct 1.76234... by 10.

Question 2

This was a well-answered question with many candidates scoring full marks in part (a)

and part (b) providing most of the discrimination.

In part (a), most candidates differentiated correctly, factorised out d

d

y

x and rearranged

their equation to arrive at the correct gradient function. Some candidates did not apply

the product rule correctly when differentiating ,xy while a few candidates left the

constant term ‘ 1 ’ as part of their differentiated equation or did not differentiate the

term 2.y

In part (b), most candidates set the numerator of their d 2 4

d 5 2

y x y

x x y

equal to 0.

Some candidates gave up at this point while others set the denominator of their d

d

y

x

equal to 0 and attempted to solve their 2 4 0x y and their 5 2 0x y

simultaneously. Many candidates substituted their 4 2y x into the curve C, but some

struggled to obtain a correct 2 2 1 0x x (or equivalent), mainly due to algebraic,

sign or bracketing errors. Those who progressed this far usually produced a correct

method for solving their 3-term quadratic. Only a minority of candidates found the

correct exact values 1 2 ,x and some of them wasted their time by needlessly

finding the corresponding values for y. A few candidates applied the alternative method

of substituting their 4

2

yx

into the curve C, finding both values of y and using their

4

2

yx

to find both values of x.

Question 3

Part (i) was generally well answered, with most candidates scoring full marks. Part (ii)

and part (iii) were discriminating with many candidates unable to score in part (ii).

In part (i), most candidates used the given 2 2

13 4

(2 1) ( 3) (2 1) (2 1) ( 3)

x A B C

x x x x x

to write down the correct identity 213 4 (2 1)( 3) ( 3) (2 1) .x A x x B x C x

The incorrect identity 213 4 (2 1)( 3) ( 3) (2 1)x A x x B x C x was used by a

few candidates. Most candidates used either a method of substituting values into their

identity or a method of comparing coefficients and were generally successful in finding

the correct values for A, B and C. Most candidates used their constants in a correct

method of integrating the resulting partial fractions, with many finding a correct 1ln(2 1) 3(2 1) ln( 3)x x x c which some simplified to give

3 3ln .

2 1 2 1

xc

x x

Common errors included integrating

2

(2 1)x

to give

2ln(2 1)x or integrating 2

6

(2 1)x to give

6

2 1x

or 23ln(2 1) .x

In part (ii), a minority of candidates expanded 3(e 1)x to give 3 2e 3e 3e 1x x x , and

many of these integrated to give a correct 3 21 3

e e 3e ,3 2

x x x x c with a few giving

the incorrect 3 23e 6e 3e .x x x x c Some candidates applied the substitution exu

to achieve the correct 2 1

3 3 d ,u u uu

with many of them integrating to achieve

a correct answer in terms of x. It was rare to see a correct solution for those candidates

who used the substitution e 1.xu Most of them achieved the correct 3

d ,( 1)

uu

u

but only a few applied a method of long division to give 2 1

1 d .1

u u uu

Some candidates did not attempt part (ii), while many gave incorrect answers such as

4 41 1(e 1) or (e 1) .

4e 4

x x

x

In part (iii), many candidates differentiated 3u x to give

2d3

d

xu

u and applied the

substitution correctly to give 2

3

3d .

4 5

uu

u u Some candidates made no further progress

while many simplified the integral to give the correct 2

3d

4 5

uu

u with some

simplifying this further to give the incorrect 3 3

d .4 5

uu

u

Those candidates who

found a correct 2

3d 1

d 3

ux

x

were less successful in obtaining a correct integral in terms

of u. Some candidates recognised that 2

3d

4 5

uu

u could be expressed in the form

f ( )d

f ( )

uu

u

and integrated to give 2ln(4 5) .u c While many of these candidates

obtained the correct 3

,8

common incorrect values of included 8 1

,3 8

or 3. Many

candidates gave their final answer in terms of u, and only a minority applied 1

3u x to

arrive at a final answer in terms of x.

Question 4

This question discriminated well between the average and more able candidates with

part (b) more accessible than part (a). There were a few candidates, however, who

made no creditable progress in this question.

In part (a), some candidates struggled to make any creditable progress. Some

candidates wrote down 3

hr (or equivalent) with no supported working, while

others used the given result to write down 3 21 1.

9 3 3

hh r h r These ploys,

which usually led to candidates arriving at the given answer, received no marks. Many

candidates used trigonometry to write down a correct equation such as tan 30,r

h

tan 60,h

r

50 tan 30

50

r

h

or ,

sin30 sin 60

r h and a few candidates applied

Pythagoras’ Theorem to give the correct equation 2 2 2(2 ) .h r r Many of these

candidates substituted either ,3

hr

3

3r h or

2 21

3r h into

21

3V r h and

arrived at the given result 31.

9V h

In part (b), most candidates differentiated to give 2d 1

d 3

Vh

h and many used the chain

rule to write down a correct equation for d

.d

h

t Many candidates divided 200 by their

d

d

V

h and substituted 15h to arrive at an exact value for

d.

d

h

t Common errors

included: applying d

200 their ;d

V

h

applying d

their 200d

V

h

and leaving their

final answer as either d 200

d 75

h

t or

d 8.

d 3

h

t

Question 5

This question discriminated well between candidates of all abilities with many

candidates struggling to access the final mark in part (a) and the final mark in part (b).

In part (a), most candidates applied a full method of setting 2y to find t and

substituting their t into 1 5sinx t t to find the exact value of k. Only a minority

of candidates found and applied 2

t

to give the correct answer 62

k

or

12.

2k

Many candidates used the incorrect value

2t

which led to the incorrect

answer 4.2

k

Only a few of these candidates realised that 42

k

was negative

and so some of them opted to use 3

2t

(which was outside the given )t to

give the incorrect 3

6 .2

k

Some candidates found both 62

k

and 42

k

with many of them identifying the correct 62

k

as their final

answer. Other common mistakes included: working in degrees; deducing 2

k

after correctly finding ;2

t

deducing 42

k

after finding 4;2

x

and not

finding k as an exact value.

In part (b), most candidates applied the complete process of using a correct method of

parametric differentiation to find d

d

y

x, using their value of t to find the gradient of the

curve at A and applying a correct straight-line method for finding the equation of the

tangent. Occasionally, sign errors were seen in the differentiation of both x and y and

a few candidates wrote down an incorrect d

5cos .d

xt

t Only a minority of candidates

used 2

t

to find a correct 4 26 2 .y x Many candidates arrived at the

incorrect tangent equation 4 18 2y x by using their 2

t

from part (a). It was

rare to see these candidates enquire why they were applying a tangent gradient of 4

when it was clear from the diagram that the gradient of the tangent to C at the point A

is negative. Other common mistakes in part (b) included: bracketing and sign errors

when finding the equation of their tangent; and working in degrees.

Question 6

This question discriminated well between candidates of all abilities.

There were a few candidates who made no creditable progress in this question. Some

of these candidates substituted 8

x

and 2y into 2

2

d

d 3cos 2

y y

x x and attempted

to find the equation of the tangent to the curve at the point , 2 .8

Many candidates separated the variables correctly and integrated 2

1

y correctly. Some

candidates did not give a correct method for integrating 2

1.

3cos 2x Common errors

included simplifying 2

1

3cos 2x to give 23sec 2 ,x

21 1cos

3 2x

or 21

cos 2 .3

x and

applying 2

1d

3cos 2x

x to give 2 1

d ln(1 cos 4 ).3(1 cos 4 ) 6

x xx

Many candidates applied the boundary conditions 2y and 8

x

to their

integrated equation which contained a constant of integration.

Only a minority of candidates rearranged a correct 1 1 1

tan 23 6

xy

to give a

correct answer in the form f ( ).y x Many candidates, applied the misconception

1 1 1B C A

A B C to give the incorrect answer 3 6cot 2 .y x

Question 7

This question discriminated well between candidates of all abilities with many

candidates scoring full marks in part (a), part (b) and part (d). Part (c) and part (e)

provided good discrimination for the more able candidates.

In part (a), many candidates found a correct (1, 1, 4),B while a few made an arithmetic

error when adding OA to AB to give the incorrect (1, 1, 0).B Common errors

included: applying OP AB to give (5, 7, 6) and finding the difference between OA

and .AB

In part (b), most candidates applied the scalar product formula between AB and AP

(or between BA and )PA with many arriving at a correct 4ˆcos( ) 2121

PAB or

4ˆcos( ) .21

PAB The most common mistake was to apply the scalar product formula

between AB and PA (or between BA and )AP which resulted in giving

4ˆcos( ) 21,21

PAB although the negative sign was dropped by some of these

candidates. A minority of candidates applied the scalar product formula between non-

relevant vectors such as OP and OB or PA and .OB A few candidates arrived at the

correct answer by applying a cosine rule method or by applying a vector product

method.

Part (c) proved challenging with many candidates unable to find the exact value for ˆsin( ).PAB Most of these candidates found an approximation for the angle PAB and

applied the formula ‘1

sin2

ab C ’ to give 1

216 56 sin(29.205...) 26.8 (3 sf ).2

A

minority of candidates used 2 ˆ1 cos ( )PAB or used a right-angled triangle method to

find the exact value of ˆsin( ).PAB Those candidates who wrote the area expression

11 4216 56 sin cos 21

2 21

only gained the method mark for applying the

correct area formula. Only a few of these candidates received full credit by simplifying

this expression to give the correct 12 5. Other common errors in this part included

applying 1 4

216 56 sin 212 21

or

21 4

216 56 1 21 .2 21

In part (d), most candidates wrote down a correct equation for the line 2 ,l while some

candidates lost the final mark for writing an expression rather than an equation for 2.l

In part (e), some candidates applied the complete process of expressing BQ in terms

of , forming and solving the equation 0BQ AP to find and substituting their

into the line 2l to find the coordinates of Q. Most candidates who applied this

process usually found the correct coordinates of Q, while only a few made manipulation

errors in solving their equation 0BQ AP to find or in evaluating the coordinates

of Q from their correct . Some candidates applied 0BQ AP to give the equation

12 6 6 30.x y z These candidates did not receive any credit until they formed an

equation in one unknown by applying, for example, 9 4 ,x 1 6y and

8 2z to their equation in x, y and z. Many candidates, however, made no progress

with part (e), with many unable to start and some attempting to solve the incorrect

equation 0.OQ AP A few candidates used a method of similar triangles with some

of them using ratios to correctly deduce 5

,4

PQ AB where

9 4

1 6 .

8 2

OQ

Only a

few of these candidates applied 5

4 to find the correct coordinates of Q.

Question 8

This was a well-answered question with part (b) providing most of the discrimination.

In part (a), most candidates applied the method of integration by parts in the correct

direction and many produced a fully correct solution. Some candidates integrated

cos4x incorrectly to give either 1

sin 4 ,4

x sin 4x or 4sin 4 .x After correctly

obtaining 1 1

sin 4 sin 4 d ,4 4

x x x x a few candidates integrated to give the incorrect

1 1sin 4 cos 4 .

4 8x x x c

In part (b), most candidates applied the volume formula 2dy x to give the correct

42

0

sin 2 d ,x x x

although a few used incorrect formulae such as 22 d ,y x

2dy x

or even d .y x Many candidates realised that a strategy of using a double-angle

formula was required with some applying incorrect formulae such as 2cos4 2sin 2 1x x and

2cos2 1 2sin 2 .x x Some candidates applied the formula

2cos4 1 2sin 2x x to give the correct 4

0

1 cos 4d .

2

xx x

Most of these

candidates integrated by using their answer to part (a) to give an expression of the form 2 sin 4 cos4 ; , , 0.Ax Bx x C x A B C Other candidates arrived at this expression

by using an integration by parts method with u x and d 1 cos 4

.d 2

v x

x

Most

candidates applied the limits of 4

and 0 correctly, with some making the mistake of

assuming the limit of 0 would give 0 when substituted into their integrated result. There

were many bracketing, manipulation and sign errors seen in this part with only a

minority of candidates using a correct method to achieve a correct exact answer

31 1.

64 16 A few candidates used the elaborate method of integrating by parts

twice, firstly with 2sin 2u x and

d

d

vx

x followed by

2u x and d

sin 4 .d

vx

x It was

rare to see these candidates obtaining full marks.

Pearson Education Limited. Registered company number 872828

with its registered office at 80 Strand, London, WC2R 0RL, United Kingdom